Timeline for On the solution of a generalized Lyapunov equation
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Aug 5, 2014 at 12:02 | comment | added | Joris Bierkens | @FedericoPoloni: You are right, I am sure you did not mean any harm. Everything there is to say is in the link. My apologies. | |
| Aug 5, 2014 at 11:58 | comment | added | Federico Poloni | @JorisBierkens I did not know of this discussion, thanks for pointing it out. On the other hand, I find your sarcastic "thank you" slightly out of line; a simple link would have been sufficient. | |
| Aug 5, 2014 at 10:04 | answer | added | Joris Bierkens | timeline score: 3 | |
| Aug 5, 2014 at 9:50 | comment | added | Dude-Ray | Ok, buddies, let's focus on the problem. I remembered that last year, Dr. Suvrit proposed a iterative method, i.e. "if ∥∑iFi\kronFi∥<1, then starting from X0=I, you can iterate Xk+1=∑iFiXkFTi+C, and converge to the unique semidefinite solution. If the operators don't satisfy this sufficient condition, then more thought is needed" | |
| Aug 5, 2014 at 9:46 | comment | added | Joris Bierkens | @FedericoPoloni: thank you. meta.mathoverflow.net/questions/410/… | |
| Aug 5, 2014 at 9:45 | history | edited | Dude-Ray | CC BY-SA 3.0 |
added 92 characters in body
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| Aug 5, 2014 at 9:38 | comment | added | Dude-Ray | Yep, you are right. I have edited moments ago. The condition I gave may not be correct. But the usual criterion you raised is for continuous case not for discrete one. The discrete form is the eigenvalues of $A$ fall into the unit circle. | |
| Aug 5, 2014 at 9:35 | history | edited | Dude-Ray | CC BY-SA 3.0 |
deleted 137 characters in body
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| Aug 5, 2014 at 8:39 | comment | added | Federico Poloni | It is unusual (and frowned upon) to thank other users in questions on this site, so I have removed a couple of sentences from your post. Back to business now: you say that it is 'clear' that $X$ is positive definite under those conditions; based on what exactly? The usual criterion for positive-definiteness in the Lyapunov case ($B=A^T$, one term) is that all the eigenvalues of $A$ have negative real part, which seem different from (and contradicting with) with your proposed one. | |
| Aug 5, 2014 at 8:33 | history | edited | Federico Poloni | CC BY-SA 3.0 |
removed thanks; tried to make the text flow a bit better
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| Aug 5, 2014 at 3:02 | review | First posts | |||
| Aug 5, 2014 at 3:08 | |||||
| Aug 5, 2014 at 2:59 | history | asked | Dude-Ray | CC BY-SA 3.0 |